Do the grouping $x^{3}-6x^{2}-x+6=\left(x^{3}-6x^{2}\right)+\left(-x+6\right)$, and factor out $x^{2}$ in the first and $-1$ in the second group.
$$x^{2}\left(x-6\right)-\left(x-6\right)$$
Factor out common term $x-6$ by using distributive property.
$$\left(x-6\right)\left(x^{2}-1\right)$$
Consider $x^{2}-1$. Rewrite $x^{2}-1$ as $x^{2}-1^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
